Re: highest bit, statically determined

[Bill Findlay <yaldnif.w@blueyonder.co.uk>]

On 29/09/2012 18:34, in article
50673111$0$9505$9b4e6d93@newsspool1.arcor-online.net, "Georg Bauhaus"
<rm.dash-bauhaus@futureapps.de> wrote:

> Is there a shorter/better way of having the compiler
> find the highest bit = 1 in a static numeric constant?
> 
> If N is such a constant, e.g. Some_Type'Last where
> Some_Type'Size = 8 and its bound are static, then
> 
>     Highest_Bit_In_Octet : constant :=
>       Natural'Max
>       (8 * Boolean'Pos (N / 2**7 > 0),
>        Natural'Max
>        (7 * Boolean'Pos (N / 2**6 > 0),
>         Natural'Max
>         (6 * Boolean'Pos (N / 2**5 > 0),
>          Natural'Max
>          (5 * Boolean'Pos (N / 2**4 > 0),
>           Natural'Max
>           (4 * Boolean'Pos (N / 2**3 > 0),
>            Natural'Max
>            (3 * Boolean'Pos (N / 2**2 > 0),
>             Natural'Max
>             (2 * Boolean'Pos (N / 2**1 > 0),
>              Natural'Max
>              (1 * Boolean'Pos (N / 2**0 > 0), 0))))))));

In my experience that sort of code, applied in non-static cases, is less
efficient than one would hope, and more obvious code works faster.

Something like the following can be readily extended to greater operand
widths:

   function first_1_bit (y : octet)
   return Natural is
      x : octet;
      r : Natural;
   begin
      if y = 0 then return 0; end if;

      if (y / 16) /= 0 then
         r := 4; x := y / 16;
      else
         r := 0; x := y;
      end if;

      if (x / 4) /= 0 then
         r := r + 2; x := x / 4;
      end if;

      if (x / 2) /= 0 then
         r := r + 1;
      end if;

      return r + 1;
   end first_1_bit;

It looks fairly inline-able, and foldable for a static value of y.

-- 
Bill Findlay
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